Explanation:
Equation -
CaCO₃ → CaO + CO₂
molecular masses of -
CaCO₃ = 40+12+16×3 = 100 g/mol
CaO = 40+16 = 56 g/mol
CO₂ = 12+16×2 = 44 g/mol
Now,
If 100 g of CaCO₃ is decomposed then it produces 56 g of CaO and 44g of CO₂
Then,
If 1 g of CaCO₃ is decomposed then it produces 56/100g of CaO and 44/100g of CO₂
Now,
Here in question, 10g of CaCO₃ is decomposed
It means, It produces
56/100 × 10 = 56/10 = 5.6 g of CaO and
44/100 × 10 = 44/10 = 4.4 g of CO₂
Now,
we get that 10 g of CaCO₃ on burning produces 4.4 g of CO₂
we need to calculate volume of 4.4 g of CO₂
- mole = mass/Gram molecular mass
Then,
mole = 4.4/44 = 0.1
Now,
volume = mole × 22.4 (in Litres)
= 0.1 × 22.4 = 2.24 L
Hence,
2.24 L of CO₂ will be produced when 10g of CaCO₃ decomposed.
Answer:
As estrelas produzem a sua energia por um mecanismo chamado fusão nuclear. Nesse processo dois elementos simples se fundem para produzir um elemento mais pesado, liberando muita energia. ... Neste caso ocorre exatamente o contrário da fusão: Átomos muito pesados (como o Urânio) são quebrados, liberando energia.
15.5% by mass is
equivalent 15.5 g urea in 100 g solution or 155 g urea in 1 kg solution. <span>
<span>we know that molality = moles solute / kg solvent
<span>moles solute = 155 g / 60 g/mol = 2.58 moles urea
</span></span></span>
Since there are 155 g
urea in 1000g solution, hence the solvent is 845 g or 0.845 kg
So:<span>
<span>molality = 2.58 / 0.845 = 3.06 m</span></span>
Hope this helps you. found this on a website